Integration by Pn Azizah

8/20/2019 Integration by Pn Azizah

1/15

4

Skill 1 : Indefinite Integral

Skill 1.1 : Integrals of xn :

1.

3 13

3 1

x x dx c

=4

4

x c

2.5 x dx 3.

9 x dx

4.3 x dx 5.

2 x dx 6. x dx

Skill 1.2 : Integral of axn :

Note : m dx mx c , c is a constant

1. 10 dx 10x + c 2. 12

dx 3.3dx

4.410 x dx 5.

34 x dx 6.

3 136 6.

3 1

x x dx c

=4

6.4

xc

=

43

2

xc

7.

1 1

8 8.1 1

x x dx c

=

2

8.2

xc

=

2

4 x c

8. 6 x dx 9. 3 x dx

10.312 x dx 11.

28 x dx 12.510 x dx


2/15

5

13.3

3

22dx x dx

x

=

3 1

2.3 1

xc

=

2

2.2

xc

=2

1c

x

14.5

5

88dx x dx

x

=

15.4

12dx

x

16.3

2

5dx

x 17.

2

3 x dx 18.

20.9 x dx

Skill 1.3 : Integral of Algebraic Terms

Note : Integrate term by term. Expand & simplify the given expression where necessary .

Example : 2(3 4 5) x x dx =

3 23 45

3 2

x x x c = x

3 – 2x2 + 5x + c

1. (6 4 ) x dx

=

2.2(12 8 1) x x dx

=

3.3( 3 2) x x dx

=

4. (3 2 ) x x dx

=

5. (2 1)(2 1) x x dx

=

6. ( 2)( 3) x x dx

=


3/15

6

7.2(3 2) x dx

=

8.2

(2 1)(2 1) x xdx

x

=

9.

2

2

6 4 xdx

x

=

10.1 1

(3 )(3 )dx x x

=

11.2

2

1( 2 3 ) x x dx

x

=

12.2(3 ) x x dx

=

Skill 1. 4 : Integral of ( ) , 1nax b dx n

E

XA

MPL

E

4 14 (3 2)

(3 2) 3(4 1)

x x dx c

=5(3 2)

15

xc

EX

AMP

LE

4

4

12

12(2 3)(2 3) dx x dx x

=312(2 3)

3. (2)

xc

=

1.

5(2 4 ) x dx 2.

4( 2) x dx

3. 23

(2 1)dx

x

4. 615

(3 5)dx

x

1( )

( ) , 1( 1)

nn ax b

ax b dx c na n


4/15

7

5.

36(2 ) x dx

6.

330(4 3 ) x dx

7.32 (1 2 )

3 x dx 8. 4

15

2( 3)dx

x

Skill 1. 5 : Determine the equation of curve from gradient function

1 Given dx

dy = 2 x + 2 and y = 6 when x = – 1,

find y in terms of x. SPM 2003, K2

dx

dy = 2 x + 2

2 2 y x dx

=22

22

x x c

y = x2 + 2x + c y = 6, x = 1 , 6 = 12 + 2(1) + c

6 = 3 + cc = 3

Hence y = x2 + 2x + 3

2 Givendx

dy = 2x + 3 and y = 4 when x = 1, find

y in terms of x.

y = x2 + 3x

3 Given

dx

dy = 4x + 1 and y = 4 when x = – 1, find

y in terms of x.

4 Given

dx

dy = 6x – 3 and y = 3 when x = 2, find y

in terms of x.


5/15

8

y = 2x2 + x + 3y = 3x2 – 3x – 3

5 Given

dx

dy = 4 – 2x and y = 5 when x = 1, find

y in terms of x.

y = 4x – x2 +2

6 Given

dx

dy = 3x2 – 2 and y = 4 when x = – 1, find

y in terms of x.

y = x3 – 2x + 3

7 The gradient function of a curve which passes

through A(1, – 12) is 3x2 – 6x . Find theequation of the curve. SPM 2004, K2

dx

dy = 3 x2 – 6x

23 6 y x x dx

=3 23 6

3 2

x xc

y = x3 – 3x2 + c

y = – 12, x = 1, – 12 = 13 – 3(1) + c

– 12 = – 2 + cc = – 10

Hence y = x3 – 6x – 10

8. The gradient function of a curve which passesthrough B(1, 5) is 3x

2 + 2 . Find the equation

of the curve.

dx

dy = 3 x2 + 2

23 2 y x dx =

y = x3 + 2x + 2

9. The gradient function of a curve which passesthrough P(1, – 3) is 4x – 6 . Find theequation of the curve.

y = 2x2 – 6x + 1

10 The gradient function of a curve which passesthrough Q( – 1 , 4) is 3x (x – 2) . Find theequation of the curve.

y = x3 –3x2 + 8


6/15

9

Skill 2 : Definite Integral

Skill 2.1 : Integral of Algebraic Terms

1.

22

2

11

2 x dx x

= 22 – 12

= 4 – 1

= 3

2.

33

3 4

00

4 x dx x

=

[81]

3.

2

2

1

6 x dx

=

[14]

4.

2

2

1

3 x dx

=

[ 3

2]

5.

3

3

1

2( ) dx x

=

[ 8

9]

6.

2

2

1

3

2dx

x

=

[ 3

4]

7.

3

0

(2 6 ) x dx

[-21]

8.

3

2

1

(4 3 ) x x dx

[-10]

9.

3

0

(2 1) x x dx

[22.5]

10.

2

1

(2 1)(2 1) x x dx

=

[ 253

]

11.

3

2

1

(3 2) x dx

=

[38]

12.

1

0

(3 2) x x dx

=

[1]


7/15

10

Skill 2.2 : Definite Integral of b

a

ndxbax )( , n ≠ –1

EXA

MP

LE

11 54

0 0

(3 2)(3 2)

5.3

x x dx

=1

5

0

(3 2)

15

x

=5 5

5 2

15 15

= 206.2

Yo

u

Tr

y!

2 2

2

21 1

33(2 1)

(2 1)dx x dx

x

=

=

1.

1

3

0

16(2 4 ) x dx

[1280]

2.

1

3

0

6( 2) x dx

[

3.

2

2

1

6

(2 1)dx

x

[2]

4.

3

3

2

24

(3 5)dx

x

[3


8/15

11

Skill 2. 3 : Applications of Definite Integral

Given that

3 3

1 1

( ) 4 , ( ) 6 . f x dx g x dx Find the value of .

1.

1

3

( ) f x dx =

-4

2. 3

1

2 ( ) f x dx =

8

3. 3

1

4 ( ) 2 f x x dx =

8

4.

3

1

3 ( ) 1

2

f xdx

=

5

5. 3

1

3 ( ) 2 ( ) f x g x dx =

0

6.

3

1

12 ( ) ( )

2 g x f x dx

=

14

7Given

2

1( ) 3 f x dx and

2

3( ) 7 f x dx .

Find

a) 3

15 ( ) 1 f x dx

(b) 2

1the value of k if ( ) 8kx f x dx

(a) 48

(b) k = 22

3

8Given that

4

0( ) 3 f x dx and

4

0( ) 5 g x dx . Find

(a)4 0

0 4( ) ( ) f x dx g x dx

(b) 4

03 ( ) ( ) f x g x dx

(a) – 15

(b) 4


9/15

12

Skill 2. 4 : SPM Questions

1. Given that

1

(2 3) 6

k

x dx

, where k > – 1, find

the value of k. SPM 2004

4

2. Given that

0

(2 1) 12

k

x dx , where k > 0, find the

value of k.

4

3. Given that

0

(3 4 ) 20

k

x dx , where k > 0, find

the value of k.

4

4. Given that

0

(6 1)

k

x dx = 14 , where k > 0, find

the value of k.

2

5. Given that6

2

( ) 7 f x and 6

2

2 ( ) 10 f x kx ,

find the value of k . SPM 2005

¼

6. Given that5

1

( ) 8 g x dx , find

(a) the value of1

5

( ) , g x dx

(b) the value of k if 5

1

( )kx g x dx = 10.

SPM 2006

(a) -8 (b)3

2


10/15

13

7. Diagram shows the curve y = f(x) cutting the

x-axis at x = a and x = b. SPM 2006

Given that the area of the shaded region is 5 unit2,

find the value of 2 ( )

b

a

f x dx .

Answer : ......................

8. Diagram 9 shows the curve y = f(x) cutting the

x-axis at x = a and x = b.

Given that the area of the shaded region is 6 unit2, find

the value of 3 ( )a

b

g x dx .

Answer : ..........................

9. Diagram shows part of the curve y = f(x).

Given that4

0

( ) f x dx = 15 unit2, find the area of

the shaded region.

Answer : ......................

10

.

Diagram shows part of the curve y = f(x).

Given that the area of the shaded region is 40 unit2,

find the value of8

0

( ) f x dx .

Answer : ......................

11. Diagram shows the sketch of part of a curve.(SPM 2001)

(a) Shade, on the given diagram, the region

represented by

8

2

x dy .

(b) Find the value of

4 8

0 2

y dx x dy

Answer : (b) ................ ......

12

.

Diagram shows the sketch of part of a curve.

(a) Shade, on the given diagram, the region represented

by10

2

x dy .

(b) If10

2

x dy = p , find , in terms of p, the value of

6

0

y dx .

Answer : (b) .............. ........

x

y

O a b

y = f(x)

x

y

O a b

y = g(x)

x

y

O

y = f(x)

5

4 x

y

O

y = f(x)

6

8

x

y

O

2

(4, 8)●

x

y

O

10

(6, 2)●


11/15

14

13. Given that 45

(1 )dx

x = k(1+ x)n + c , find

the value of k and n. SPM20’03

[ 5 , 33

k n ]

14

.

Given that3

12

(3 2)dx

x = k(3x – 2 )n + c , find the

value of k and n.

k = – 2 , n = – 2

15. Given that 1

2 2

0

416 10

3 x kx k dx .

Find the possible values of k. (SPM 2002)

k = – 1 , -4

16

.

Given that 1

2

0

3 10 4 0 x kx dx . Find the value

of k.

k = – 1

17. (SPM 01) Given that2

( )1

d x g x

dx x

, find

the value of 3

2

2 ( ) x g x dx .

.

9

2

18 Given that ( )1

d x f x

dx x

, find the value of

3

2

4 ( ) x f x dx .

.

113

2


12/15

15

Skill 3 : Find the Area of the region between a curve and the axes

1. Calculate the area of the shaded region

.Given y = 2x2

+x3

37 1/3 unit2

2. Find the area enclosed by the curve

y = x(x – 1)(x – 2)

1/2 unit2

3. Find the area of the shaded region

9 unit2

4 Find the area of the shaded region

4 1/2 unit2

Homework : Text Book – Exercise 3.2.2 page 72 no 17

'

Area =

b y

a y

xdy Area=

b x

a x

ydx

The shaded area between the curvey =f (x) , x = a , x = b and the x -axis

x

5

2

0

x =y 2 – 7y +10

The shaded area between the curvey = f (x ), y =a , y = b and the y -axis


13/15

16

Skill 4 : Find the Volume of Revolution

Calculate the volume generated

when the shaded region is revolved 3600 about the x-axis

π

5

26

π

15

1113 π20

111

y = x +1

y = x(x – 2)

The volume of revolution V, generated when thearea under a curve y = f(x) by x-axis from x = a tox = b is rotated about the x-axis

dx y I b

a

x 2

The volume of revolution V, generated when thearea under a curve y = f(x) by x-axis from x = a tox = b is rotated about the y-axis

dy x I b

a

y 2

π

3

226

π

15

11


14/15

17

Homework : Text Book – Exercise 3.2.2 page 72 no 18 and 19

1.

SPM 1993

Calculate the volume of the solid generatedwhen the shaded region in the diagram isrevolved through 360

o about the y-axis

2

1

2.

SPM 1998

Diagram below shows the graph of y = x2 – 2 and

straight line 153

y x

Calculate the volume of the

solid generated when the shaded region in thediagram is revolved through 360

o about the y-axis

3 π π

2

140

π

1 π

3

31

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00 aabboouutt tthhee yy--aaxxiiss

13π

y = x2 – 2

y = x2+ 1

153

y x


15/15

18

Integration by Pn Azizah

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